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The Computable Differential Equation Lecture ... - Bruce E. Shapiro

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108 CHAPTER 5. RUNGE-KUTTA METHODS<br />

Hence the iteration formula (equation 5.155) gives<br />

y n = y n−1 + hf(y n−1 ) + 1 2 h2 f y f + 1 6 h3 (f 2 y f + f yy f 2 ) + O(h 4 ) (5.162)<br />

But since y ′ = f, y ′′ = f y f, and y ′′′ = f 2 y f 2 + f yy f 2 , the exact expansion is identical.<br />

Thus the method is at least O(h 3 ).(In fact the error is precisely O(h 3 ) but this is<br />

left as an exercise).<br />

Thus in contrast to explicit methods, implicit methods may have order higher<br />

than the number of stages; we shall see that they may, in fact have an order as high<br />

as twice the number of stages. <strong>The</strong> price we pay is that the method is implicit –<br />

some additional work, such as a Newton iteration – is required to implement the<br />

method.<br />

Figure 5.6: Region of absolute stability for the method of equation 5.153. <strong>The</strong> region<br />

of absolute stability is the region outside the indicated curve.<br />

Implicit methods can be absolutely stable in a large fraction of the negative real<br />

plane, making them potentially good stiff solvers. For example, the method 5.153<br />

when applied to the test equation gives<br />

Solving the pair of equations for K 1 and K 2 gives<br />

K 1 = y n−1 λ + h 4 (K 1 − K 2 )λ (5.163)<br />

K 2 = y n−1 λ + h 12 (3K 1 + 5K 2 )λ (5.164)<br />

(5.165)<br />

6λ − 4hλ2<br />

K 1 =<br />

h 2 λ 2 − 4hλ + 6 y n−1 (5.166)<br />

6λ<br />

K 2 =<br />

h 2 λ 2 − 4hλ + 6 y n−1 (5.167)<br />

Math 582B, Spring 2007<br />

California State University Northridge<br />

c○2007, B.E.<strong>Shapiro</strong><br />

Last revised: May 23, 2007

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